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3 years ago

14

[4] A tortoise and a hare cover the same distance in a race. The hare goes very fast but stops frequently while the tortoise has

a steady pace and finish first

1 answer:

nadya68 [22]3 years ago

7 0

Answer:

I know that story where the hare sleeps

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3 Below, someone is trying to balance a plank with

belka [17]

Answer:

a. The moment of the 4 N force is 16 N·m clockwise

b. The moment of the 6 N force is 12 N·m anticlockwise

Explanation:

In the figure, we have;

The distance from the point 'O', to the 6 N force = 2 m

The position of the 6 N force relative to the point 'O' = To the left of 'O'

The distance from the point 'O', to the 4 N force = 4 m

The position of the 4 N force relative to the point 'O' = To the right of 'O'

a. The moment of a force about a point, M = The force, F × The perpendicular distance of the force from the point

a. The moment of the 4 N force = 4 N × 4 m = 16 N·m clockwise

b. The moment of the 6 N force = 6 N × 2 m = 12 N·m anticlockwise.

8 0

3 years ago

A kilogram is a unit of mass and when multiplied by meters per second

amm1812

Answer:

it yields a derived unit known as Newton, which is the unit of Force.

Explanation:

The formula for force is given by Newton's Second Law, which states that whenever an unbalanced force is applied to a body, it produces an acceleration in the body in the direction of force.

$F = ma\\$

where, F = Force

m = mass

a = acceleration

Now, we substitute the respective S.I units of each quantity in equation:

F = Newton (N)

m = kilogram (kg)

a = acceleration = m/s²

Therefore,

$N = (kg)(m/s^2)$

So, it is clear from above expression that:

<u>When kilogram is multiplied by meter per second squared, it yields a derived unit known as Newton, which is the unit of Force.</u>

5 0

3 years ago

2-The amount of internal energy needed to raise the temperature of 0.25kg of water by 0.2°C is 209.3 J. How fast must a 0.25 kg

Yakvenalex [24]

Answer:

40.92 m/s

Explanation:

The computation is shown below:

Ek = 1 ÷2mv²...............................(1)

v = √(2Ek/m).......................... (2)

Here EK denotes kinetic energy

m denotes mass

v denotes velocity

Given that

m = 0.25kg and Ek = 209.3J

So,

v = √(2×209.3 ÷0.25)

= √1674.4

= 40.92 m/s

7 0

3 years ago

(a) You wish to determine the height of the smokestack of a local coal burning power plant. You convince a member of the mainten

damaskus [11]

SHİNE LİKE A BRİGHT STAR

❄NEVER GİVE UP❄

☻NEVER SAY NEVER☻

4 0

3 years ago

Consider an object with s=12cm that produces an image with s′=15cm. Note that whenever you are working with a physical object, t

Leni [432]

A. 6.67 cm

The focal length of the lens can be found by using the lens equation:

$\frac{1}{f}=\frac{1}{s}+\frac{1}{s'}$

where we have

f = focal length

s = 12 cm is the distance of the object from the lens

s' = 15 cm is the distance of the image from the lens

Solving the equation for f, we find

$\frac{1}{f}=\frac{1}{12 cm}+\frac{1}{15 cm}=0.15 cm^{-1}\\f=\frac{1}{0.15 cm^{-1}}=6.67 cm$

B. Converging

According to sign convention for lenses, we have:

- Converging (convex) lenses have focal length with positive sign

- Diverging (concave) lenses have focal length with negative sign

In this case, the focal length of the lens is positive, so the lens is a converging lens.

C. -1.25

The magnification of the lens is given by

$M=-\frac{s'}{s}$

where

s' = 15 cm is the distance of the image from the lens

s = 12 cm is the distance of the object from the lens

Substituting into the equation, we find

$M=-\frac{15 cm}{12 cm}=-1.25$

D. Real and inverted

The magnification equation can be also rewritten as

$M=\frac{y'}{y}$

where

y' is the size of the image

y is the size of the object

Re-arranging it, we have

$y'=My$

Since in this case M is negative, it means that y' has opposite sign compared to y: this means that the image is inverted.

Also, the sign of s' tells us if the image is real of virtual. In fact:

- s' is positive: image is real

- s' is negative: image is virtual

In this case, s' is positive, so the image is real.

E. Virtual

In this case, the magnification is 5/9, so we have

$M=\frac{5}{9}=-\frac{s'}{s}$

which can be rewritten as

$s'=-M s = -\frac{5}{9}s$

which means that s' has opposite sign than s: therefore, the image is virtual.

F. 12.0 cm

From the magnification equation, we can write

$s'=-Ms$

and then we can substitute it into the lens equation:

$\frac{1}{f}=\frac{1}{s}+\frac{1}{s'}\\\frac{1}{f}=\frac{1}{s}+\frac{1}{-Ms}$

and we can solve for s:

$\frac{1}{f}=\frac{M-1}{Ms}\\f=\frac{Ms}{M-1}\\s=\frac{f(M-1)}{M}=\frac{(-15 cm)(\frac{5}{9}-1}{\frac{5}{9}}=12.0 cm$

G. -6.67 cm

Now the image distance can be directly found by using again the magnification equation:

$s'=-Ms=-\frac{5}{9}(12.0 cm)=-6.67 cm$

And the sign of s' (negative) also tells us that the image is virtual.

H. -24.0 cm

In this case, the image is twice as tall as the object, so the magnification is

M = 2

and the distance of the image from the lens is

s' = -24 cm

The problem is asking us for the image distance: however, this is already given by the problem,

s' = -24 cm

so, this is the answer. And the fact that its sign is negative tells us that the image is virtual.

3 0

4 years ago